In a committee of 5, which must be selected from 4 males and 3 females. In how many ways can the members be chosen if it were to include 2 females?
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Correct Answer: Option D
Explanation:
For the committee to include 2 females, we must have 3 males, so that there should be 5 members.
That is, \(^4C_3 \times ^3C_2\)
= \(\frac{4!}{(4 - 3)! 3!} \times \frac{3!}{(3 - 2)! 2!}\)
= 4 × 3 = 12 ways
For the committee to include 2 females, we must have 3 males, so that there should be 5 members.
That is, \(^4C_3 \times ^3C_2\)
= \(\frac{4!}{(4 - 3)! 3!} \times \frac{3!}{(3 - 2)! 2!}\)
= 4 × 3 = 12 ways